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Math Help - How many eggs?

  1. #1
    A riddle wrapped in an enigma
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    How many eggs?

    Once upon a time, and old lady went to sell her vast quantity of eggs at the local market.

    When asked how many she had, she replied:

    Son, I can't count past 100 but I know that.

    If you divide the number of eggs by 2 there will be one egg left.
    If you divide the number of eggs by 3 there will be one egg left.
    If you divide the number of eggs by 4 there will be one egg left.
    If you divide the number of eggs by 5 there will be one egg left.
    If you divide the number of eggs by 6 there will be one egg left.
    If you divide the number of eggs by 7 there will be one egg left.
    If you divide the number of eggs by 8 there will be one egg left.
    If you divide the number of eggs by 9 there will be one egg left.
    If you divide the number of eggs by 10 there will be one egg left.

    Finally. If you divide the number of eggs by 11 there will be NO EGGS left!

    How many eggs did the old lady have?

    HINT #1
    Spoiler:
    Find a number X into which all of the numbers from 2 to 10 divide evenly. You can do this by simply using 2*3*4*5*6*7*8*9*10, but you can find a smaller number by finding the prime factors, a subset of which can be used to form any number from 2 to 10. 2*2*2*3*3*5*7 will do. This comes out to be 2520, and is the lowest number into which all the numbers 2-10 divide evenly.


    HINT #2
    Spoiler:
    We can add 1 to this number to satisfy the first 9 constraints of the puzzle (the remainder of 2521/2, 2521/3 ... 2521/10 is one), but this does not satisfy the last constraint, divisibility by 11.


    ANSWER
    Spoiler:

    Fortunately, we can multiply X (=2520) by any integer and add 1 and we will still satisfy constraints 1-9. So what Y do we multiply X by so that (X*Y) + 1 is divisible by 11. 2520/11 has a remainder of 1. Thus two 2520s divided by eleven would have a remainder of 1+1 = 2, and so forth...so ten 2520s divided by 11 would have a remainder of 10. This number plus one would divide eleven evenly, as well as also satisfy the first 9 constraints - thus 25201 is the answer.
    Last edited by masters; May 18th 2009 at 10:04 AM.
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by masters View Post
    Once upon a time, and old lady went to sell her vast quantity of eggs at the local market.

    When asked how many she had, she replied:

    Son, I can't count past 100 but I know that.

    If you divide the number of eggs by 2 there will be one egg left.
    If you divide the number of eggs by 3 there will be one egg left.
    If you divide the number of eggs by 4 there will be one egg left.
    If you divide the number of eggs by 5 there will be one egg left.
    If you divide the number of eggs by 6 there will be one egg left.
    If you divide the number of eggs by 7 there will be one egg left.
    If you divide the number of eggs by 8 there will be one egg left.
    If you divide the number of eggs by 9 there will be one egg left.
    If you divide the number of eggs by 10 there will be one egg left.

    Finally. If you divide the number of eggs by 11 there will be NO EGGS left!

    How many eggs did the old lady have?
    Hi masters.

    The minimum number of eggs is 25201, but she could also have had 52921, 80641, 108361,

    In general:

    \fbox{$27720n-2519$} for all n\in\mathbb Z^+
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  3. #3
    MHF Contributor
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    We know that x = 11a where a \in \mathbb{N}.

    Also, all number's  x that when divided by 2, 3, \cdots 10 that leaves 1 as a remainder are of the form

    x = 2520 b + 1 where b \in \mathbb{N}.

    This sets up a diophantine equation

     <br />
11a - 2520b = 1<br />

    whose solution is

    a = 2520 k - 229,\;;\ b = 11k - 1

    so x = 2520(11k-1) + 1 = 27720k - 2519 as previous stated with the minimum being when k = 1.

    Just thought I'd put some details in.
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  4. #4
    A riddle wrapped in an enigma
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    To Danny Arrigo and TheAbstractionist:

    Both of you are correct. I suppose I should've been clearer about choosing the minimum amount, but both of you had excellent responses.
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